A sequence of reactions is shown below starting from an optically active compound P. $$\Large{\ce{\underset{\ce{C4H8O3}}{P}->[SOCl2]\underset{\ce{C4H6Cl2O}}{Q}->[C2H5NH2]\underset{\ce{C6H12ClNO}}{R}}}$$ P does not react with 2,4-dinitrophenylhydrazine (2,4-DNP) to form a hydrazone. Q reacts very rapidly with one molar equivalent of ethylamine at low temperature to form R. Select the most appropriate statement(s). (A) Q as well as R are optically active. (B) Q on heating with an excess of ethylamine in the presence of a base gives a basic compound. (C) R is a basic compound and can react with an acid to form a salt. (D) An optically inactive isomer of P on heating can form compound S $(\ce{C4H6O2})$ which does not react with thionyl chloride.

My approach:

P is hydroxy butanoic acid. In Q the hydroxyl groups gets replaced by chlorine. I have confusion about the structure of R.

I feel Option (A) to be correct. I am doubtful about (D). And have no idea about (B) and (C).

• Some hint: 1. Why you think there is optical activity involved? It is not mentioned in the question; 2. Check the thionyl chloride reaction to find out how hydroxy carboxylic acid will react with it. It is more than you have thought; 3. Amine is basic but amide is not; 4. Alkyl chloride will not react with amine easily as room temperature, you need more reactive specie to react with amine "very rapidly". May 12, 2014 at 16:01

• @Rudstar Like Ron already suggested, the optically inactive isomer of P is 4-hydroxybutanoic acid, because it does not contain a stereocenter. On heating, this compound can form $\gamma$-butyrolactone by intramolecular esterification, which would be compound S. This lactone does not react with thionyl chloride. May 12, 2014 at 18:44